Integral formulation for the solution of $xy'' + y' = y$ Let's say that $y$ satisfies the following ODE:
$$xy'' + y' = y$$ 
I want to formulate $y$ as a contour integral.
I know that the final result I should get is:
$$y(x)=\frac{1}{2i\pi} \int_{C}{\frac{1}{t}e^{\sqrt{x}(t+1/t)}dt}$$
where $C$ is an appropriate contour. 
However, I don't know where to start in order to derive this result.
PS: There are some similarities between this integral formulation and that of the Modified Bessel Function of the First Kind, $I_n(x)$, as we can observe by looking at expression (1) in Mathworld.
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With $\ds{x = -\,{1 \over 4}\,t^{2}\quad\imp\quad t = 2\ic x^{1/2}}$:

\begin{align}
\totald{}{x}&=\totald{t}{x}\,\totald{}{t}=\ic x^{-1/2}\,\totald{}{t}
=-\,{2 \over t}\,\totald{}{t}
\\[3mm]\totald[2]{}{x}&=\underbrace{-\,{2 \over t}}_{\ds{\totald{t}{x}}}\
\totald{}{t}\bracks{-\,{2 \over t}\,\totald{}{t}}
={4 \over t^{2}}\,\totald[2]{}{t} - {4 \over t^{3}}\,\totald{}{t}
\end{align}

Set $\ds{{\rm u}\pars{t} \equiv {\rm y}\pars{-\,{1 \over 4}\,t^{2}}}$:
$$
-\,{1 \over 4}\,t^{2}\bracks{%
{4 \over t^{2}}\,
\totald[2]{{\rm u}\pars{t}}{t} - {4 \over t^{3}}\,\totald{{\rm u}\pars{t}}{t}}
- {2 \over t}\,\totald{{\rm u}\pars{t}}{t} = {\rm u}\pars{t}
$$

and
  $$
t^{2}\,\totald[2]{{\rm u}\pars{t}}{t} + t\,\totald{{\rm u}\pars{t}}{t} +
\pars{t^{2} - 0^{2}}{\rm u}\pars{t} = 0
$$
  which is a Bessel Equation.
  Solutions are given for linear combinations of $\ds{{\rm u}\pars{2\ic\root{x}}}$
  where $\ds{{\rm u}}$'s are Bessel functions. 

$\color{#c00000}{\ds{\mbox{The integral representation should arise from the Bessel functions integral representations.}}}$
A: Let $y=\int_Ce^{xs}K(s)~ds$ ,
Then $x(\int_Ce^{xs}K(s)~ds)''+(\int_Ce^{xs}K(s)~ds)'-\int_Ce^{xs}K(s)~ds=0$
$x\int_Cs^2e^{xs}K(s)~ds+\int_Cse^{xs}K(s)~ds-\int_Ce^{xs}K(s)~ds=0$
$\int_Cs^2e^{xs}K(s)~d(xs)+\int_C(s-1)e^{xs}K(s)~ds=0$
$\int_Cs^2K(s)~d(e^{xs})+\int_C(s-1)e^{xs}K(s)~ds=0$
$[s^2e^{xs}K(s)]_C-\int_Ce^{xs}~d(s^2K(s))+\int_C(s-1)e^{xs}K(s)~ds=0$
$[s^2e^{xs}K(s)]_C-\int_Ce^{xs}(s^2K'(s)+2sK(s))~ds+\int_C(s-1)e^{xs}K(s)~ds=0$
$[s^2e^{xs}K(s)]_C-\int_Ce^{xs}(s^2K'(s)+(s+1)K(s))~ds=0$
$\therefore s^2K'(s)+(s+1)K(s)=0$
$s^2K'(s)=-(s+1)K(s)$
$\dfrac{K'(s)}{K(s)}=-\dfrac{1}{s}-\dfrac{1}{s^2}$
$\int\dfrac{K'(s)}{K(s)}ds=-\int\left(\dfrac{1}{s}+\dfrac{1}{s^2}\right)ds$
$\ln K(s)=-\ln s+\dfrac{1}{s}+c_1$
$K(s)=\dfrac{ce^\frac{1}{s}}{s}$
$\therefore y=\int_C\dfrac{ce^{xs+\frac{1}{s}}}{s}~ds$
But since the above procedure in fact suitable for any complex number $s$ ,
$\therefore y_n=\int_{a_n}^{b_n}\dfrac{c_ne^{xk_nt+\frac{1}{k_nt}}}{k_nt}d(k_nt)=c_n\int_{a_n}^{b_n}\dfrac{e^{k_nxt+\frac{1}{k_nt}}}{t}dt$
For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $k_n$ such that:
$\lim\limits_{t\to a_n}te^{k_nxt+\frac{1}{k_nt}}=\lim\limits_{t\to b_n}te^{k_nxt+\frac{1}{k_nt}}$
$\int_{a_n}^{b_n}\dfrac{e^{k_nxt+\frac{1}{k_nt}}}{t}dt$ converges
For $n=1$ , the best choice is $a_1=0$ , $b_1=\infty$ , $k_1=-1$ when $\text{Re}(x)\geq0$
$\therefore y_1=C_1\int_0^\infty\dfrac{e^{-xt-\frac{1}{t}}}{t}dt$ when $\text{Re}(x)\geq0$
A: Assume: $$y(x) = \int_C \! e^{xt}f(t)\,\mathrm{d}t$$
This means that:
$$y'(x) = \int_C \! te^{xt}f(t)\,\mathrm{d}t,\,\, \text{and} \,\,y''(x) = \int_C \! t^2e^{xt}f(t)\,\mathrm{d}t.$$
Substitute this into our equation to find:
$$\int_C  (xt^2+t-1)e^{xt}f(t)\,\mathrm{d}t = 0$$
We are going to find another function $g(t)$ such that it has the following properties:
$$g(t) = t^2f(t) \,\,\, \text{and} \, \,\, g'(t) = (t-1)f(t)$$
With clever manipulations (aka division) we say that $$\log(g) = \int\!\dfrac{g'(t)}{g(t)}\mathrm{d}t = \int\! \dfrac{t-1}{t^2}\mathrm{d}t = \log(t) + \dfrac{1}{t}$$
$$g(t) = te^{1/t}$$
$$f(t) = \dfrac{1}{t}e^{1/t}$$
Now we can say that our solution is 
$$y(x) = \int_C \! \frac{1}{t}e^{xt+1/t}\,\mathrm{d}t.$$
$g(t)$ was chosen in such a way that $e^{xt}(xg(t) + g'(t)) = (xt^2+t-1)e^{xt}f(t) = (e^{xt}t^2f(t))'$.
This is the technique I learned to solve equations of your type. I would imagine that assuming a form like $y(x) = \int_c \!e^{\sqrt{x}t}f(t)\mathrm{d}t$ would result in a solution similar to the one you have provided. Naturally, the factor of $1/(2\pi i)$ may be added at the end.
